package com.inuker.solution;

import java.util.HashSet;

/**
 * Created by dingjikerbo on 17/1/2.
 */

/**
 * https://discuss.leetcode.com/topic/56052/really-easy-understanding-solution-o-n-java
 * http://www.cnblogs.com/grandyang/p/5825619.html
 */
public class PerfectRectangle {

    /**
     * 这题思路是扫描所有点，满足这两个条件就能构成完美矩形，
     * 1. 所有矩形的顶点中，除了四个角只能出现一次，其余的点必须出现偶数次
     * 2. 所有矩形的面积和必须等于四个角的点构成的面积，为了防止重叠
     */
    public boolean isRectangleCover(int[][] rectangles) {
        if (rectangles.length == 0 || rectangles[0].length == 0) {
            return false;
        }

        int x1 = Integer.MAX_VALUE;
        int x2 = Integer.MIN_VALUE;
        int y1 = Integer.MAX_VALUE;
        int y2 = Integer.MIN_VALUE;

        HashSet<String> set = new HashSet<String>();
        int area = 0;

        for (int[] rect : rectangles) {
            x1 = Math.min(rect[0], x1);
            y1 = Math.min(rect[1], y1);
            x2 = Math.max(rect[2], x2);
            y2 = Math.max(rect[3], y2);

            area += (rect[2] - rect[0]) * (rect[3] - rect[1]);

            String s1 = rect[0] + " " + rect[1];
            String s2 = rect[0] + " " + rect[3];
            String s3 = rect[2] + " " + rect[3];
            String s4 = rect[2] + " " + rect[1];

            if (!set.add(s1)) set.remove(s1);
            if (!set.add(s2)) set.remove(s2);
            if (!set.add(s3)) set.remove(s3);
            if (!set.add(s4)) set.remove(s4);
        }

        if (!set.contains(x1 + " " + y1) || !set.contains(x1 + " " + y2) || !set.contains(x2 + " " + y1) || !set.contains(x2 + " " + y2)) {
            return false;
        }

        return set.size() == 4 && area == (x2 - x1) * (y2 - y1);
    }
}
